1958 AMC 12 Problem 42

Attempt Problem 42 of the 1958 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1958 AMC 12 solutions, or check the answer key.

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42.

In a circle with center

O,

chord

\overline{AB}

equals chord

\overline{AC}.

Chord

\overline{AD}

cuts

\overline{BC}

in

E.

If

AC=12

and

AE=8,

then

AD

equals:

27

24

21

20

18

Answer: E

Concepts:chord inscribed angle similarity

Difficulty rating: 1790

Small hint:

Compare triangles

AEC

and

ACD

Big hint:

Angles

ACE

and

ADC

subtend the equal chords

AB

and

AC

Solution:

Because

E

lies on

BC

and

AD,

triangles

AEC

and

ACD

share angle

A.

Also

\angle ACE=\angle ACB

subtends chord

AB,

while

\angle ADC

subtends chord

AC.

Since

AB=AC,

these angles are equal. Thus

\triangle AEC\sim\triangle ACD.

Corresponding sides give

\frac{AE}{AC}=\frac{AC}{AD}.

Therefore

\frac{8}{12}=\frac{12}{AD},

so

AD=18.

Thus, the correct answer is E.

Problem 42 in other years